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Equivalent conductance of saturated ${BaSO}_{4}$ is 400 ${ohm}^{- 1} {cm}^{2} {equiv}^{- 1}$ and specific conductance is $8 x 10^{- 5} {ohm}^{- 1}, {cm}^{- 1 .}$ . Hence $K_{sp} of {BaSO}_{4}$ is

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$1 x 10^{- 4} M^{2}$

$2 x 10^{- 4} M^{2}$

$1 x 10^{- 8} M^{2}$

$4 x 10^{- 8} M^{2}$

answer is B.

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Detailed Solution

𝜆 = 400

K = 8 × $10^{- 5}$

$N = \frac{8 X 10^{- 5} X 10^{3}}{400} = 2 X 10^{- 4}$

$S = M = \frac{N}{2} = 1 X 10^{- 4}$

$∴ k_{SP} = S^{2} = {(10^{- 4})}^{2} = 10^{- 8}$

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